Problem 38 - Astronomy: At Play in the Cosmos

Question

Radiation from a nearby star has heated a cloud in the ISM from \(100 \mathrm{K}\) to \(550 \mathrm{K},\) and its density has dropped to 0.25 times its earlier density. What is the ratio of its current pressure to its pressure before these changes?

Step-by-Step Solution

Answer

The pressure increased by a factor of 1.375.

Step 1: Understand the Relation of Pressure, Temperature, and Density

In a given system, pressure \( P \) is related to temperature \( T \) and density \( \rho \) by the ideal gas law: \( P = n k T \), where \( n \) is the number density and \( k \) is the Boltzmann constant. If the density changes to a fraction of its original, the new density can be expressed as a multiple of the original density.

Step 2: Set Up the Ideal Gas Law for Initial and Final Conditions

Initially, let the pressure be \( P_1 \), the temperature be \( T_1 = 100 \text{ K} \), and the density be \( \rho_1 \). Finally, after the change, let the pressure be \( P_2 \), the temperature \( T_2 = 550 \text{ K} \), and the density \( \rho_2 = 0.25 \rho_1 \).

Step 3: Express Pressure in Terms of Known Quantities

Using the ideal gas law in terms of density, we have \( P = \rho k T \). Therefore, the initial pressure \( P_1 = \rho_1 k T_1 \) and the final pressure \( P_2 = \rho_2 k T_2 \).

Step 4: Calculate the Ratio of Final to Initial Pressure

The ratio of the final pressure to the initial pressure is \( \frac{P_2}{P_1} = \frac{\rho_2 k T_2}{\rho_1 k T_1} = \frac{\rho_2 T_2}{\rho_1 T_1} \). Substitute \( \rho_2 = 0.25 \rho_1 \), \( T_2 = 550 \text{ K} \), and \( T_1 = 100 \text{ K} \) into the equation:\[ \frac{P_2}{P_1} = \frac{0.25 \rho_1 \times 550}{\rho_1 \times 100} = \frac{0.25 \times 550}{100} \].

Step 5: Simplify the Expression

Now, simplify the expression:\[ \frac{0.25 \times 550}{100} = \frac{137.5}{100} = 1.375 \].

Step 6: Interpret the Result

The ratio \( \frac{P_2}{P_1} = 1.375 \). This means the current pressure is 1.375 times the initial pressure.

Key Concepts

Pressure in the Ideal Gas Law

Pressure is a fundamental concept in the context of gases, referring to the force that the gas exerts against the walls of its container. In the ideal gas law, we relate pressure to other parameters such as temperature and density. The formula is expressed as \( P = n k T \), where:

• \( P \) stands for pressure.
• \( n \) represents the number density, which is essentially a measure of the number of molecules per unit volume.
• \( k \) is the Boltzmann constant, which helps relate energy at the single-molecule level to temperature.
• \( T \) is the absolute temperature of the gas in Kelvin.

Pressure conveniently allows us to understand how a gas behaves under various conditions by indicating how compact or energetic the gas molecules are. In the provided exercise, a change in temperature and density affects the pressure, as shown through the calculated ratio of pressures before and after heating the gas cloud.

Temperature and Its Role

Temperature provides a measure of the average kinetic energy of the molecules in a gas. In the context of the ideal gas law, it plays a crucial role by directly influencing the pressure. When the temperature of a gas increases, the energy of its molecules rises, causing them to move more vigorously. This increased movement translates to a greater force or pressure exerted on the walls of the container.
In the given exercise, the cloud's temperature increased from 100 K to 550 K due to radiation from a star. This significant rise in temperature would naturally lead to an increase in pressure, provided the number of molecules remains constant. As seen from the relation \( P = n k T \), pressure is directly proportional to temperature, meaning if one increases, so does the other, given constant density settings.

Understanding Density in Gases

Density in the context of gases refers to the mass per unit volume, which can often be linked to the concentration of gas molecules within that volume. In the ideal gas law, density can directly influence pressure when considered with a constant temperature. High density implies more molecules are present in a given space, potentially creating more pressure due to more frequent molecular collisions.
The exercise in question illustrates a scenario where the density drops to 0.25 its previous value. This decrease reflects fewer gas molecules per unit volume, potentially leading to reduced pressure. However, with the rise in temperature factored in, we see a complex interplay, where the reduction in density is offset by the increase in temperature, ultimately influencing the ratio of pressures calculated. Here, an understanding of density helps explain how changes in external conditions like heating from a star can affect both density and overall gas behavior in terms of pressure.